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Puzzle: Banter

Greg says, “I’ll tell Phil the product of two integers, each of which is greater than one, and Sean their sum.”
After Greg whispers to both of them, Phil shrugs, “I don’t know Greg’s numbers.”
“I already knew that,” Sean states.
“But now I do!” Phil exclaims.
Sean smirks, “Well then so do I.”

19 Comments

Sean already knows the product P isn’t sufficient to identify the numbers x and y such that P = xy because he knows that P doesn’t have a unique factorization into two numbers. He must know this because he knows that P has at least three factors. The only way to know this for certain is if he received a number S that was odd. Then he knows that x is odd and y is even hence xy has at least three factors. (If S were even then it could be the sum of two odd numbers so it could be the sum of two odd primes and P would only have two factors.)

Now Phil knows the sum of x and y is odd and therefore x is odd and y is even. He knows x and y now because among the 2-factorizations of his number P there is only one with one odd number and one even number, and some 2-factorizations into two even numbers. Call this Phil’s property. (These numbers must have the form 2 * 2 * p for some odd prime p.)

Now Sean knows that Phil’s number only has one 2-factorization containing one odd number and one even number, and some 2-factorizations into two even numbers. He knows x and y now because among the 2-partitions of his number S there is only one pair whose product satisfies Phil’s property.

The numbers could be 4 and 3. Then P = 12 and S = 7. But there are other potential numbers such as 4 and 5. Then P = 20 and S = 9. etc.

You’re right, x = 3, y = 2 would be a counterexample. I clearly glossed over all the base cases. x=3, y=2 is sort of a base case. If S = 5, then Sean wouldn’t have said “I already knew Phil didn’t know the number” because for S = 5, there is only one 2-partition having x, y > 1 which is x = 3, y =2, which makes P = 6 which has only two factors and Sean would know that Phil would know the factors immediately.

As far as my intuition, yes, my statement was too strong that “the only way he could know for certain…” is by way of the parity. I definitely made a leap that the information is gleaned from the parity. I explained why the parity seems to work in my post. (If S were even, then it could be the sum of two odd _numbers_, and I guess I assumed that would allow for some odd primes, in which case Sean wouldn’t be able to say Phil couldn’t determine the two factors of P. Maybe my assumption there was the leap.)

I suppose one could search for a theorem that says “Any number N satisfying T has 2-partitions where every 2-partition has at least one composite number.” But that seems way too hard. I can see how this is related to Goldbach’s conjecture but at the time I didn’t think Goldbach’s conjecture was necessary (maybe it is now that I’m thinking about it).

Again, I could be totally wrong and there’s probably some massively simplifying thing I’m missing like in the last problem which I solved at home in a very roundabout way without applying the pigeonhole principle.

OK I’m struggling here but I think you need two conditions on S then to work around the counterexamples (like x = 2, and y = any odd prime e.g. S = 9 = 2 + 7 is bad).

I think you need:
– S is odd
– S – 2 is composite

Then all the 2-partitions (x,y) x, y > 1 of S break down as:
– odd + even, i.e. x is odd and y is even (without loss of generality)
– then y is either 2 or some other even number > 2
– If y > 2 then x and y have three factors altogether
– If y = 2, then since S – 2 is composite we have x = S – y and x is composite so again we have three factors altogether between x and y

Going back to the original question we can choose two numbers x, y, S = x + y, P = xy, such that S is odd, S – 2 is composite. Also x must be an odd prime and y must be a power of two.

Like x = 7, y = 4, S = 11, P = 28. Phil says I don’t know x and y. Then Sean says I know you didn’t know by the earlier proof. Phil says Ah, x and y must be odd and even. He now picks out 7 and 4 as the only pair out of { (14, 2), (7, 4) } with odd parity. So Phil says now I know. Then Sean says ah, you only have one odd parity pair in your 2-factorization. So your numbers must be of the form (odd prime, power of two). Only one 2-partition of S satisfies that, x = 7 and y = 4.

x = 13, y = 4, S = 17, P = 52. Phil says I don’t know x and y there are too many factors. Then Sean says I know you didn’t know by the earlier proof. Phil says Ah, S must be odd so x and y are odd and even. He picks out (4, 13) as the only pair out of {(26, 2), (13, 4)} with odd parity. So Phil says now I know. Then Sean says ah, you only have one odd parity pair in your 2-factorization so your numbers must be of the form (power of two, odd prime). In my 2-partitions, I have { (2, 15), (4, 13), (8, 9) } containing powers of 2 so it must be (4, 13).

Oh ok. So basically choose a number that can’t be expressed as the sum of two primes. Then Sean knows Phil can’t determine the factors immediately. Then Phil knows it must be odd (by Goldbachs conjecture) so one number must be even and the other odd. Then the rest of what I said.

Choose numbers S, x, y, S = x + y and S can’t be expressed as the sum of two prime numbers. Sean says I know you don’t know the numbers because S can’t be expressed as the sum of two prime numbers.

Then Phil says Ah you know this because your number can’t be expressed as the sum of two prime numbers so (assuming Goldbach’s conjecture) your number is odd. Hence one of (x, y) must be odd and the other even. Looking at P, Phil only sees one 2-factorization satisfying that and now knows the number.

Then Sean says Ah, P only has one 2-factorization into two numbers odd * even. Hence, P must be of the form P = odd prime * 2^k. Looking through the 2-partitions of S he only finds one pair (a, b) where a is an odd prime times a power of two and the other is a power of two.

Take x = 4, y= 13, S = 17, P = 52. Phil says I don’t know x, y. Sean notices 17 can’t be expressed as the sum of two primes and says I already knew that. Phil says Ah, S must be odd so x and y must be one odd and one even. Phil looks at the 2-factorizations of 52 = { (26, 2), (13, 4) } and deduces it must be (13, 4) and says I know (x, y). Sean says ah you only have one 2-factorization with odd parity so P = odd prime * power of two. Sean looks through the 2-partitions of S = 17 containing powers of 2 { (2, 15), (4, 13), (8, 9) } and figures it must be (4, 13).

“Looking through the 2-partitions of S he only finds one pair (a, b) where a is an odd prime times a power of two and the other is a power of two.”

What would happen if both a and b were even though?

“I think there’s a small epistemic gap near the beginning.”

Well… assuming Goldbach’s conjecture seems epistemologically problematic :P. The main reason I phrased the question “What might Greg’s integers be?” like you pointed out earlier, was to make the question about showing existence rather than proving uniqueness of a solution.